When $20 g$ of naphthoic acid ( $C_{1} 1 H_{8} O_{2}$ ) is dissolved in $50 g$ of benzene $(K_{f}$ = $1.72 K k g m o l^{- 1})$ freezing point depression of $2 K$ is observed. The van't Hoff factor $(i)$ is:

0.5

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Solution

The correct option is C $0.5$
Molaloity $= \frac{20 \times 1000}{172 \times 50} = 2.32$
$∴$ Depression in freezing point $= 2 = 2.32 \times 1.72 \times i$
$∴ i = 0.5$

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QUESTION 2.33

19.5 g of $C H_{2} F C O O H$ is dissolved in 500g of water. The depression in the freezing point of water observed in 1.0 C. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid.
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When 20g of naphthoic acid (C11H8O2) is dissolved in 50g of benzene (Kf=1.72Kkgmol−1) freezing point depression of 2K is observed. The van't Hoff factor (i) is:

The correct option is C 0.5Molaloity =20×1000172×50=2.32∴ Depression in freezing point =2=2.32×1.72×i∴i=0.5

When $20 g$ of naphthoic acid ( $C_{1} 1 H_{8} O_{2}$ ) is dissolved in $50 g$ of benzene $(K_{f}$ = $1.72 K k g m o l^{- 1})$ freezing point depression of $2 K$ is observed. The van't Hoff factor $(i)$ is:

The correct option is C $0.5$
Molaloity $= \frac{20 \times 1000}{172 \times 50} = 2.32$
$∴$ Depression in freezing point $= 2 = 2.32 \times 1.72 \times i$
$∴ i = 0.5$